y=k(x-1)-k^2/4=kx-(4k+k^2)/4,代入y=ax^2+bx+c(a>o)得:
kx-(4k+k^2)/4=ax^2+bx+c
ax^2+(b-k)x+(4c+4k+k^2)/4 = 0
只有一个公共点,则判别式=0,即:
(b-k)^2 - 4a(4c+4k+k^2)/4 = b^2-2bk+k^2-4ac-4ak-ak^2=0
y=k(x-1)-k^2/4=kx-(4k+k^2)/4,代入y=ax^2+bx+c(a>o)得:
kx-(4k+k^2)/4=ax^2+bx+c
ax^2+(b-k)x+(4c+4k+k^2)/4 = 0
只有一个公共点,则判别式=0,即:
(b-k)^2 - 4a(4c+4k+k^2)/4 = b^2-2bk+k^2-4ac-4ak-ak^2=0